Vinayprajapati wrote:
IF K is an integer greater than 6, then all the following must be divisible by 3 EXCEPT
a) k(k+3)(k-1)
b)3k^3
c)(k+1)(k+5)(k+6)
d)(k+2)(k-2)(k+3)
e)k(k+1)(k+2)
Since we are working with VARIABLE, We can be sure if the product can be shown as 3 consecutive numbers, by adding /subtracting 3 from any of the terms..
lets see the choices
a) k(k+3)(k-1)(k-1)*k*(k+1) are consecutives BUT instead of k+1, we have k+3 in k(k+3)(k-1) ..
and if k+1 is div by 3, k+3 will not and k and k-1 will also be not div by 3..
so this is NOT necessary to be div by 3
b)3k^33 is there in the term, so TRUE
c)(k+1)(k+5)(k+6)if K+1 is div by 3, k+1+3=k+4 will also be div by 3
so we have consecutive terms by adding 3 to k+1
k+4, k+5 and k+6..
TRUE
d)(k+2)(k-2)(k+3)add 3 to k-2. it becomes k-2+3=k+1..
again 3 consecutive terms k+1, k+2 and k+3
TRUE
e)k(k+1)(k+2)3 consecutive terms already
TRUE
When should one use plug in method to solve a question? Is this method of plugging in better?